A proton and an `alpha`-particle are accelerated through same potential difference. Find the ratio of their de-Brogile wavelength.
A
2
B
1
C
`2 sqrt 2`
D
None of these
Text Solution
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The correct Answer is:
C
The gain in K.E. of a cahnge particle after moving through a potential difference of V is given as eV, that is equal to `(1)/(2)mv^(2)` where v is the velocity of the charge particle . `(1)/(2)mv^(2) =qV implies v = sqrt ((2qV)/(m))` therefore mv= `sqrt(2mqV)`, `implies` de -Broglie wavlength ` lambda =(h)/(mv)= (h)/(sqrt(2mqV))`, `therefore` ` (lambda_(p))/(lambda_(alpha))=sqrt ((m_(alpha)q_(alpha)V_(alpha))/(m_(p)q_(p)V_(p)`, Putting `V_(alpha)=V_(p)`, `(lambda_(p))/(lambda_(alpha))= sqrt(((4)(2))/((1)(1)))=2sqrt2`
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